## Permutation and Combination Quiz-9

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background.

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced.

Q1.How many words can be formed from the letters of the word ARTICLE, if vowels always comes at the odd places?
•  60
•  576
•  7!/3!
•  120
Solution

(b) In a word ARTICLE, there are 7 letters. Out of 7 places, 4 places are odd and 3 even.
Therefore 3 vowels can be arranged in 4 odd places in 4 P3 ways and remaining 4 consonants can be arranged in 4 P4 ways.
Hence, required number of ways = 4 P3×4 P4=576
Q2.Five digited numbers with distinct digits are formed by using the digits, 5, 4, 3, 2, 1, 0. The number of those numbers which are multiples of 3, is
•  720
•  240
•  216
•  120
Solution
(c) A number is divisible by 3, if the sum of the digits is divisible by 3
Since, 1+2+3+4+5=15 is divisible by 3, therefore total such numbers is 5! ie, 120
And, other five digits whose sum is divisible by 3 are 0, 1, 2, 4, 5
Therefore, number of such formed numbers=5!-4!=96
Hence, the required number if numbers=120+96=216

Q3. There are n seats round a table numbered 1,2,3,…,n. The number of ways in which m(≤n) persons can take seat is
•  nCm
•  nCm.m!
•  (m-1)!
•  (m-1)!*(n-1)!
Solution
(b) As the seats are numbered so the arrangement is not circular
Hence, required number of arrangements = nCm.m!

Q4. If the letters of the word ‘SACHIN’ are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number
•  72
•  70
•  69
•  71
Solution
(b) Factorizing the given number, we have 38808=23×32×72×11
The total number of divisors of this number is same as the number of ways of selecting some or all of two 2' s, two 3’s, two 7’s and one 11.
Therefore, The total number of divisors =(3+1)(2+1)(1+1)-1 =71
But, this includes the division by the number itself
Hence, the required number of divisors =71-1=70

Q5.How many 10 digits numbers can be written by using digits 9 and 2?
•  10C1*9C2
•  210
•  10C2
•  10!
Solution
(b) Total number of numbers=2×2×2…10 times=210

Q6.Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
•  9
•  12
• 10
•  14
Solution
(b) Let the total number of persons in the room =n
∴ Total number of handshakes=nC2=66 (given)
⇒ n!/(2!(n-2)!)=66
⇒(n(n-1))/2=66
⇒ n2-n-132=0
⇒ (n-12)(n+11)=0
⇒ n=12 [∵n≠-11]

Q7.If (2n+1) P(n-1): (2n-1) Pn:3:5, then the value of n is equal to
•  4
•  3
•  2
•  1
Solution

Q8.The number of ways in which a committee can be formed of 5 members from 6 men and 4 women if the committee has at least one woman, is
•  366000
•  660000
•  360000
•  3999960
Solution

=360((100000-1)/9)=40×99999=3999960

Q9.The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines, is
•  6
•  18
•  12
•  9
Solution
(c) From the first set, the number of ways of selection two lines =4 C2
From the second set, the number of ways of selection two lines =3C2
Since, these sets are intersect, therefore they from a parallelogram,
∴ Required number of ways = 4C2 ×3C2 =4×3=12

Q10.

•  56C4
•  56C3
•  55C4
• 55C4
Solution

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